an arithmetical progression has positive terms .the ratio of the difference of the 4th and 8th term to the 15th term is 4/15 and square of the difference of the 4th and the 1st term is 255.which term of the series is 2015??
options
1)225 2)404 3)403 4)410
please answer
I think that the problem is supposed to say that "square of the difference of the 4th and the 1st term is 225" rather than 255......when I tried 255 I get a strange square root which doesn't allow for any "integer" answers [If I didn't make a mistake, that is !!!!].....I'm working this one assuming 225 is correct...
The first term is just a1
Let the 4th term be given by:
a4 = a1 + d(4 -1) = a1 + 3d
And the difference between the 4th and 1st terms is just 3d
And the square of this difference = 225
So (3d)^2 = 225 → 9d^2 = 225 → d^2 = 225/9 → d = 15/3 so d =5 [since all the terms are positive]
And we're told that [ (a1 + d(7) ) - (a1 + d(3)) ] / [a1 + d (14) ] = 4/15 simplify
[d(7) - d(3)] /[ a1 + d(14)] = 4/15 and sustituting 5 for d we have
[5(7) - 5(3)] /[ a1 + 5(14)] = 4/15
5*4 = [4/15] (a1 + (5)*14) multiply through by 15
5*4*15 = 4[a1 + 70]
300 = 4a1 + 280
20 = 4(a1)
So a1 = 5
So
2015 = 5 + 5(n -1)
2010 = 5(n -1)
402 = n - 1 add 1 to both sides
403 which is answer "3"
I think that the problem is supposed to say that "square of the difference of the 4th and the 1st term is 225" rather than 255......when I tried 255 I get a strange square root which doesn't allow for any "integer" answers [If I didn't make a mistake, that is !!!!].....I'm working this one assuming 225 is correct...
The first term is just a1
Let the 4th term be given by:
a4 = a1 + d(4 -1) = a1 + 3d
And the difference between the 4th and 1st terms is just 3d
And the square of this difference = 225
So (3d)^2 = 225 → 9d^2 = 225 → d^2 = 225/9 → d = 15/3 so d =5 [since all the terms are positive]
And we're told that [ (a1 + d(7) ) - (a1 + d(3)) ] / [a1 + d (14) ] = 4/15 simplify
[d(7) - d(3)] /[ a1 + d(14)] = 4/15 and sustituting 5 for d we have
[5(7) - 5(3)] /[ a1 + 5(14)] = 4/15
5*4 = [4/15] (a1 + (5)*14) multiply through by 15
5*4*15 = 4[a1 + 70]
300 = 4a1 + 280
20 = 4(a1)
So a1 = 5
So
2015 = 5 + 5(n -1)
2010 = 5(n -1)
402 = n - 1 add 1 to both sides
403 which is answer "3"